Lecture 3
Reduction of D-(+)-Camphor
Introduction
• Reduction of Ketones and Aldehydes
Reactant
Ketone
Product
Alkane
Ketone
Ketone
Ketone
Aldehyde
Ketone
Alcohol
Alcohol
Alcohol
Alcohol + Acid
Diol
Reagent
Name
Zn/HCl
N2H4/KOH
H2/Ni
Al(OCH(CH3)2)3
NaBH4, LiAlH4
KOH
Mg-metal
hn/(CH3)2CHOH
Clemmensen
Wolff-Kishner
Raney Nickel
Meerwein-Ponndorf
Cannizzaro
Pinacol
Pinacol for aromatic
ketones
Mechanism I
• In Chem 30BL, sodium borohydride (NaBH4) will be used
as the reducing agent
• Driving force for reaction is the formation of a very strong
B-O bond vs. the p-bond of the carbonyl group and the
B-H bond
• The light elements of group 3 often form compounds that
possess a partial double bond character in the E-X bond,
if X has one or more lone pairs i.e., N, O, F, etc.
Mechanism II
–
H BH3
R2C O
H BH3
R2C O
O
3R C R
(occurs 3 mo re times)
Keton e o r Aldeh yde
–
R2 CHO B(OCHR2 )3
OH
4 R C R
H
H O CH3
+ (CH3O)4B –Na+
(large ex cess)

(deco mp oses at
R
H C R
H
O
R
–
R C O B O C R
R
O
H
H C R
R
"tetraalkyl b orate"
elevated temp eratures)
• Ultimately, two hydrogen atoms are added to the ketone:
one originates from the hydride (:H-), which forms the
C-H function, and the other one from the protic solvent
(H+) that leads to the formation of the hydroxyl function
Mechanism (Stereochemistry I)
• The reduction of 2-pentanone affords a racemic mixture of 2-pentanol
because the activation energies (G‡) for the two alternate pathways are
identical
• The reduction of D-(+)-camphor affords a mixture of two diastereomeric
alcohols. The exo product (=(-)-isoborneol) is formed in larger quantity
compared to the endo product (=(+)-borneol) because the activation energy
for the formation of the exo product is lower
Mechanism (Stereochemistry II)
• The stereochemistry of the reaction can be explained using
HOMO-LUMO concept
– The hydride is the nucleophile in the reaction which provides
the electrons for the newly formed C-H bond
– The carbonyl group is the electrophile in the reaction and
therefore has to provide an empty orbital for the reaction
(p*(C=O), LUMO)
p
p-bond
p-bond
exo approach
C
100-120o
C
O
C
O
O
p
endo-approach
HOMO of a C=O bond
(side view)
LUMO of a C=O group
(side view)
The p-orbital is closer to the oxygen atom,
hence the oxygen contributes more to the
orbital (bigger lobes). In the p*-orbital the
situation reverses.
Mechanism (Stereochemistry III)
•
Exo approach
camphor
•
Endo approach
camphor
Exo approach
2-norbornanone
Bottom line:
– Camphor: exo approach is sterically more hindered resulting in a higher
activation energy for this pathway and a lower quantity of the endo
product (=borneol)
– 2-norbornanone: the exo approach is less hindered resulting in the endo
product as major product
Mechanism (Stereochemistry IV)
• The stereoselectivity for the reaction would be higher
– if the R-group on the side of the carbonyl function was
increased in size
– if the size of the nucleophile was increased
Reducing agent
NaBH4
LiAlH4
LiAlH(OMe)3
LiBH(n-Bu)3
LiBH(sec-Bu)3
LiBH(iso-amyl)3
2-Norbornanone
(endo product)
86 %
89 %
98 %
98 %
99.6 %
>99.5 %
Camphor
(exo product)
86 %
92 %
99 %
98 %
99 %
99.3 %
– if the reaction temperature was lowered
Experimental Design
•
Choice of Reducing Agent
– LiAlH4 : causes a high stereoselectivity, more reactive (can even be pyrophoric);
requires very dry diethyl ether or very dry tetrahydrofuran as a solvent 
– NaBH4 : much safer but strong enough of a reducing reagent to reduce the ketone 
•
Choice of Solvent
– NaBH4: moderately soluble in water, insoluble in diethyl ether
– Camphor: very poorly soluble in water (0.1 g/100 mL), well soluble in diethyl ether
– The solvent choice is a compromise in terms of polarity: methanol dissolves both
compounds reasonably well (NaBH4: 13 g/100 mL, camphor: 63.1 g/100 mL)
– Problem: Sodium borohydride reacts with protic solvents 
– Solution
• A large excess of the reducing agent is used to ensure the complete reduction of
the camphor
• Camphor is dissolved in a small amount of methanol before the NaBH 4 is added,
which takes advantage of the fact the reduction of the camphor is faster than
hydrolysis of NaBH4
Experiment I
• Dissolve the camphor in a small
amount of methanol in a 25 mL
Erlenmeyer flask
• Add the sodium borohydride in three
portions
Watch glass with ice
Boiling stick of
appropriate length
• Bring the suspension to a gentle boil
• After the reaction is completed, place
the solution in a cold water bath
• What is the setup here?
• Why?
• Add ice-cold water to the reaction
mixture
• Isolate the solid using vacuum
filtration
• Why is water added?
• Suck air through the solid for at least
10 minutes
• Why is air sucked through the
solid?
To have better control over the reaction
during the addition of the water
To complete the hydrolysis and
to precipitate the organic compounds
(1.2 mg/mL borneol in water at 25 oC)
To remove the bulk of the water from
the solid
Experiment II
• Dissolve the solid in a small
amount of diethyl ether
• Add a small amount of drying
agent (MgSO4)
• Why is the solid dissolved again?
• Remove the drying agent
• Extract the drying agent with a
small amount of diethyl ether
• How is accomplished?
• Why is this step necessary?
In order to dry it
• What is the student looking for here?
1. Some free floating drying agent
2. A transparent solution
To recover some of the adsorbed product
• Why is the drying agent removed?
1. The drying process is reversible
2. The product and the drying agents
are both white solids which makes it
impossible to separate them later
• Remove the solvent using the
rotary evaporator
• Why is the rotary evaporator used?
• How this piece of equipment work?
See video for details
Characterization I
• Melting point (~ 1 mm in melting point capillary)
– Too much sample will result in a broader melting point range
• Infrared spectrum
• Isoborneol (KBr):
– n(OH)=3398 cm-1
(broad peak)
– n(C-OH)=1069 cm-1
(strong)
– n(C=O)=1744 cm-1
is absent!
n(C=O)
n(OH)
Isoborneol
n(C-OH)
Borneol
n(C-OH)
• Borneol (KBr):
– n(OH)=3352 cm-1
(broad peak)
– n(C-OH)=1055 cm-1
(strong)
n(OH)
Characterization II
• Gas chromatography
– Prepare a solution of the final product in diethyl ether
(conc: ~1 mg/mL)
– Fill the GC vial to the 1.5 mL mark
– Close the vial with a cap and submit into tray
– The sample cannot contain any undissolved
solids or water because they will cause significant
problems during the data acquisition
– Sign the sample in on the sign-in sheet: student name
and code on the vial (make sure not to remove it).
Do not forget to record the code in your notebook
as well.
– Samples that are not signed in will not be run!
– Pick up the printout in YH 3077E during the afternoon
of the next day
Polarimetry I
• Optical activity was discovered by E.L. Malus (1808)
• Chiral molecules rotate the plane of polarization of
polarized light
Polarizer
Analyte
Analyzer
• How does it work?
– Monochromatic light is polarized by a Nicol prism (polarizer)
– The plane-polarized light passes through a polarimetry cell in
which the plane of the light will be rotated if the cells contains
a chiral compound (or a mixture of chiral compounds)
– The analyzer rotates the plane of the light back to its original
orientation
Polarimetry II
• The value of the optical rotation (a) of a sample depends on
the wavelength (the subscript “D” refers to l=589.3 nm), the
path length (l), the concentration (c) and the specific optical
rotation for the specific enantiomer and to a lesser degree on
the temperature (X)
  []X
D *c*l
• The sign of the optical rotation is independent from the
absolute configuration!
• The sign and absolute value can depend on the solvent because
the observer might look at different compounds i.e., cation,
anion or neutral specie for amino acids.
Polarimetry III
• Polarimeter (located in YH 1096 for Chem 30BL)
• Concentration: ~1 % in 95 % ethanol (the exact concentration in g/mL has
to be known)
• It is important that there are no air bubbles in the path of the light because
they will cause problems in the measurement (i.e., dark sample error)
• The ratio of (-)-isoborneol and (+)-borneol can be calculated by
•
=x(-34.6o)+(1-x)(+37.7o)
=specific optical rotation of the sample after concentration correction
x =the mole fraction of isoborneol in the sample
[]D= +37.7o for (+)-borneol and []D= -34.6o for (-)-isoborneol
Polarimetry IV
• What influences the result in the polarimetry
measurement?
– The concentration of the sample
– A wet sample will yield a less negative value because
the concentration is less than assumed, which results in
a lower reading for the sample
– The presence of unreacted camphor ([]D= +44.26o )
– The ratio of the (-)-isoborneol and (+)-borneol i.e., a
80:20 mixture should result in a value of []= ~ -20o
after the concentration correction